Problem: Consider the polar curve $r=4\sin(3\theta)$. What is the slope of the tangent line to the curve $r$ when $\theta=\dfrac{\pi}{3}$ ? Give an exact expression. $\text{slope }=$
Solution: The slope of the tangent line at a point is equal to $\dfrac{dy}{dx}$ at that point. In the case of polar curves, we can use the relationship: $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)}$ For a polar curve, $x={r}\cos(\theta)$ and $y={r}\sin(\theta)$. Therefore, in our problem we have: $\begin{aligned} x&={4\sin(3\theta)}\cos(\theta) \\\\ y&={4\sin(3\theta)} \sin(\theta) \end{aligned}$ Let's find $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)} \\\\ &=\dfrac{12\cos(3\theta)\sin(\theta)+4\sin(3\theta)\cos(\theta)}{12\cos(3\theta)\cos(\theta)-4\sin(3\theta)\sin(\theta)} \\\\ &=\dfrac{3\cos(3\theta)\sin(\theta)+\sin(3\theta)\cos(\theta)}{3\cos(3\theta)\cos(\theta)-\sin(3\theta)\sin(\theta)} \end{aligned}$ Finally, we evaluate $\dfrac{dy}{dx}$ at ${\theta = \dfrac{\pi}{3}}$. $\begin{aligned} &\phantom{=}\left.\dfrac{dy}{dx}\right|_{{\tfrac{\pi}{3}}} \\\\ &=\dfrac{3\cos\left(3\left({\dfrac{\pi}{3}}\right)\right)\sin\left({\dfrac{\pi}{3}}\right)+\sin\left(3\left({\dfrac{\pi}{3}}\right)\right)\cos\left({\dfrac{\pi}{3}}\right)}{3\cos\left(3\left({\dfrac{\pi}{3}}\right)\right)\cos\left({\dfrac{\pi}{3}}\right)-\sin\left(3\left({\dfrac{\pi}{3}}\right)\right)\sin\left({\dfrac{\pi}{3}}\right)} \\\\ &=\dfrac{3\cos\left(\pi\right)\sin\left(\dfrac{\pi}{3}\right)+\sin\left(\pi\right)\cos\left(\dfrac{\pi}{3}\right)}{3\cos\left(\pi\right)\cos\left(\dfrac{\pi}{3}\right)-\sin\left(\pi\right)\sin\left(\dfrac{\pi}{3}\right)} \\\\ &=\dfrac{3(-1)\left(\dfrac{\sqrt 3}{2}\right)+(0)\left(\dfrac12\right)}{3(-1)\left(\dfrac12\right)-(0)\left(\dfrac{\sqrt 3}{2}\right)} \\\\ &=\sqrt{3} \end{aligned}$ The slope of the tangent line to the curve $r$ when $\theta=\dfrac{\pi}{3}$ equals $\sqrt{3}$. The graph of the tangent is shown. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${0}$ ${\frac{1}{6}\pi}$ ${\frac{1}{3}\pi}$ ${\frac{1}{2}\pi}$ ${\frac{2}{3}\pi}$ ${\frac{5}{6}\pi}$ ${\pi}$ ${\frac{7}{6}\pi}$ ${\frac{4}{3}\pi}$ ${\frac{3}{2}\pi}$ ${\frac{5}{3}\pi}$ ${\frac{11}{6}\pi}$